# Proving Convergence of a Series

## Homework Statement

($$\sqrt{(n+1)}$$ - $$\sqrt{n}$$ ) $$/$$ $$\sqrt{n}$$
I'm trying to show this series converges.

## Homework Equations

divergence test: as n approaches infinity, if the sequence does not approach 0 then the series diverges
ratio test: as n approaches infinity, if the ratio between subsequent terms is less than 1 then the series converges, if equal to 1 the test is inconclusive, if greater than 1 then the series diverges
comparison test: if this series is smaller than another series that converges, then this series also converges; if this series is greater than another series that diverges, then this series also diverges

## The Attempt at a Solution

Applying the divergence test, the sequence definitely approaches 0, but this cannot be used to prove convergence
I tried using ratio test (which has worked so well for nearly every question) and ended up with a mess of square roots.
I'm thinking of using comparison test, but what series can I compare it to?

hunt_mat
Homework Helper
Note that:
$$1=n+1-n=(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})$$
and hence
$$\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$$

Sorry but I don't see where you're going with this, can you explain?

He wants you to rationalise the numerator.

I know what he did there but I don't know how that step leads to proving convergence

hunt_mat
Homework Helper
Ideally I would like you to compare your series to the harmonic series which is known to diverge

yes hunt, but this series is smaller than the harmonic series, thus you can't compare it to the harmonic to prove divergence, maybe it's more plausible to compare it to 1/2n, and use the limit comparison test instead?

yes hunt, but this series is smaller than the harmonic series, thus you can't compare it to the harmonic to prove divergence, maybe it's more plausible to compare it to 1/2n, and use the limit comparison test instead?

$$2 \sqrt{n+1} > \sqrt{n} + \sqrt{n+1}$$

$$\frac{1}{ 2 \sqrt{n+1}} < \frac{1}{ \sqrt{n} + \sqrt{n+1}}$$

$$\frac{1}{2(n+1)} < \frac{1}{ 2 \sqrt{n+1}}$$

The series diverges.